Trial 25.58mL Volume of NaOH added to reach the equivalence point : Volume of NaOH added to reach the 1/2 equivalence point: 12.79mL 4.45 PH @ 1/₂ the equivalence point: pha of unknown acid: ≈ 4.45 Average pha of unknown Acid: 4.6 Possible Identity of unknown weak acid: Molarity of unknown acid: Trial 2 25.32mL 12.66mL 4.75 ~4.75

Trial 25.58mL Volume of NaOH added to reach the equivalence point : Volume of NaOH added to reach the 1/2 equivalence point: 12.79mL 4.45 PH @ 1/₂ the equivalence point: pha of unknown acid: ≈ 4.45 Average pha of unknown Acid: 4.6 Possible Identity of unknown weak acid: Molarity of unknown acid: Trial 2 25.32mL 12.66mL 4.75 ~4.75

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter16: Applications Of Neutralization Titrations

Section: Chapter Questions

Problem 16.20QAP

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Based on the analysis of a vinegar solution via titration with 0.1 M NaOH. Results are as follows: Trial 1 Volume of vinegar solution used (mL) 5.00 Final burette reading (mL) 44.20 Initial burette reading (mL) 0.00 Determine the percentage mass of acetic acid in vinegar for Trial 1. The density of vinegar is 1.01 g/mL and molecular mass for acetic acid is 60 g/mol.
1. 1093-g sample of impure Na2CO3 was analyzed by residual precipitimetry. After adding 50.00 mL of 0.06911 M AgNO3, the sample was back-titrated with 0.05781 M KSCN, requiring 27.36 mL to reach the endpoint. The percentage Na2CO3 (MW = 106.0 g/mole) in the tested sample is ________ % ? Note: Express final answer using least number of significant figures. 2. The alkalinity of natural waters is usually controlled by OH- (MW = 17.01 g/mole), CO3-2 (MW = 60.01 g/mole), and HCO3- (MW = 61.01 g/mole), which may be present singularly or in combination. Titrating a 10.0-mL sample to a phenolphthalein endpoint requires 38.12 mL of a 0.5812 M solution of HCl, and an additional 18.67 mL of the same titrant to reach the methyl orange endpoint. The composition of the sample is _________% CO3-2 and ___________ % OH- Note: Express final answers using least number of significant figures.
A local company sent you their green alternative for window cleaner to be tested for percent (w/v) acetic acid content. For your experiment, you first standardized your NaOH titrant with 0.8053 g of (99.80 % purity) KHP. You used 40.60 mL of NaOH for your standardization. After that you then analyzed a 10.00 mL sample and found that you needed 43.20 mL NaOH to reach the end point. Summary of results: Standardization Sample analysis KHP Weight (g) 0.8053 g Volume of sample 50.00 mL Purity 99.80% NaOH (mL) used 33.20 mL NaOH (mL) used 40.60 mL Determine the following: Molarity of NaOH % (w/v) acetic acid
Sources of Error Determine the relationship between the observed/apparent value (EX) VERSUS that of the true value (ET) for the quantity being sought by writing either <, >, or = on the space provided TOPIC: Measured mass of the precipitate 1. Filter paper was dried prior to filtration. EX _____ ET TOPIC: Standardization of Titrant 2. Distilled water was not equilibrated to room temperature before the preparation of NaOH titrant. EX ______ ET TOPIC: Determination of Molar Concentration of each component (Double Indicator Titration) 3. No blank correction EX ______ ET
PS. Further values required for the solvings are give in the various situations below. (ANSWER) Situation: A community in a mountainous area of Bohol uses water collected from a nearby natural spring. A sample was submitted to a laboratory for the analysis of its total hardness. Required: SHOW YOUR COMPLETE CALCULATIONS. BASED ON THE IMAGE PROVIDED BELOW FOR THIS QUESTION: Calculate the amount of titrant used in each trial to reach endpoint. Report total hardness of the sample as mean ±sd. a. 250.0 mL of 500.0 ppm of CaCO3 solution from a primary standard (assume solvent is distilled water only). Answer : Mass of CaCO3 = 0.125 g b, The EDTA solution was standardized by titrating it with a 25.0 mL aliquot of the CaCO3 solution. How much of the titrant was consumed. Answer: Volume of EDTA consumed = 12.405 g c. Calculate the average titer (mg CaCO3/mL EDTA). Mass of CaCO3 in 25 mL CaCO3 solution: 0.0125 g Answer: 1.008 mg CaCO3/ mL EDTA
A commercial vinegar was analyzed by titration to determine the percent acetic acid. Briefly, 10.00 mL of vinegar sample was diluted to 100. mL solution in volumetric flask. A 25.00 mL aliquot from the diluted vinegar required 25.55 mL of 0.1005 M NaOH to reach the phenolphthalein endpoint. Which is the correct equation between the analyte and titrant reaction? CH3COOH + NaOH → NaCH3COO + H2O 2CH3COOH + NaOH → NaCH3COO + H2O C20H14O4 + NaOH → NaC20H14O4 + H2O CH3COOH + 2NaOH → NaCH3COO + H2O
2,5 ml volume has taken from an “hypothetic” solution which includes (3+) Sb and (3+) Fe and at the titration with 0.1004 N KMnO4, the wasted amount has found as 16,4 mL. The other 2,5 mL that has taken, has reduced with Zn after that, this 2,5 mL solution has titrates with the same KMnO4 solution solution and the wasted amount is 26,5mL. With these datas find the %concentrations of the ions at the solution.
A RbOH solution is titrated four (4) times against potassium hydrogen phthalate (KHP; FW=204.224) samples to the Phenolphthalein endpoint. Using the data below, determine the concentration of the RbOH solution? g of KHP Volume of Base Required 0.5373 g 42.49 mL 0.5856 g 43.88 mL 0.5790 g 48.56 mL 0.5856 g 44.60 mL (Report your answer as "mean +/- std dev") M What is the percent relative standard deviation? % What is the 99% Confidence Interval for the concentration of the solution (population mean)?
Aspirin powder = 0.8110g MW of Aspirin = 180g.mol-1 Volume of 0.5N HCl consumed in back titration = 23.50mL Volume of 0.5N HCl consumed in blank titration = 44.50mL Percent purity (USP/NF) = Aspirin tablets contain NLT 90.0% and NMT 110.0% of the labeled amount of aspirin (C9H8O4) What is the calculated weight (in grams) of pure aspirin?..
A sample is analyzed for chloride by the Volhard method. From the following data, calculate the percentage of chloride present:Weight of sample = 6.0000 g dissolved and diluted to 200 mLAliquot used = 25.00 mL AgNO3 added = 40.00ml of 0.1234MKSCN for back titration = 13.20ml of 0.0930M
The buret was filled with 0.100 M HCl solution. Then was transferred in a 25.0 mL of saturated calcium hydroxide solution (2g of calcium hydroxide per 100 ml of water) in two separate E-flasks. Then 2 drops of phenolphthalein was added to each flask Titration data for the determination of solubility and Ksp of calcium hydroxide: Trial 1: Final Buret reading (ml)-19.80; Initial Buret reading (ml)- 13.00; Temperature (Celcius)- 25 Trial 2: Final Buret reading (ml)-26.10; Initial Buret reading (ml)- 19.80; Temperature (Celcius)- 25 Voume of HCl used: Trial 1- 6.80ml; Trial 2- 6.30mL 1. Compute for the moles of H+ used and the moles of OH- present. moles of H+ used = (concentration of HCl) × (volume of HCl used)moles of OH- = moles of H+ used 2. Construct an ICE table for the reaction.3. Calculate the molar solubility (in mol/L) of OH- and Ca2+.4. Determine the solubility of Ca(OH)2 in g/L. (MM of Ca(OH)2 = 74.096 g/mol). 5.…
If 500mL of 0.10M Ca2+ is mixed with 500mL of 0.10M SO42-, what mass of calcium sulfate will precipitate? Ksp for CaSO4 is 2.40 • 10-5. express answer to 3 significant figures. And include units.